# Thread: Logarithm and Exponent Question

1. I'm confused on this question.

The equation m log p (n) = q can be written in exponential form as..
The answer on the work sheet is p^(q/m)=n but shouldn't it be P^(qm) = n ? According to the power rule? My teacher explained this by writing down for me log p (n) = q / m but I'm confused here

2. Originally Posted by zekea
I'm confused on this question.

The equation m log p (n) = q can be written in exponential form as..
The answer on the work sheet is p^(q/m)=n but shouldn't it be P^(qm) = n ? According to the power rule? My teacher explained this by writing down for me log p (n) = q / m but I'm confused here
The definition of the logarithm function id the following:
If $b$ is any number such that $b>0$ and $b\neq 1$ and $x>0$ then,
$$y=\log_b x \ \ \text{ is equivalent to } \ \ b^y=x$$

We have the the equation $q=m\log_p n$.

Dividing both sides by $m$ we get $$\frac{q}{m}=\frac{m\log_p n}{m} \Rightarrow \frac{q}{m}=\log_p n$$

Therefore from the definition for $y=\frac{q}{m}$, $b=p$ and $x=n$ we get $$p^{\frac{q}{m}}=n$$

Okay this is what my teacher did but something is confusing me.

Based on Khans' video here
Basically according to the power rule you have Log a (c)^d = bd . He brought the d down to the other side.
So in exp form A^(bd) = C^d so shouldn't the answer be p^(mn) = n rather than P^(q/m) = n ?

4. Using that rule we have the following:
$$q=m\log_p n\Rightarrow q=\log_p n^m$$

Then from the definition we get $p^q=n^m$.

To solve for $n$ we do the following: $$n^m=p^q \Rightarrow \left (n^m\right )^{\frac{1}{m}}=\left (p^q\right )^{\frac{1}{m}} \Rightarrow n^{\frac{m}{m}}=p^{\frac{q}{m}} \Rightarrow n=p^{\frac{q}{m}}$$

5. Originally Posted by zekea
Okay this is what my teacher did but something is confusing me.

Based on Khans' video here
Basically according to the power rule you have Log a (c)^d = bd . He brought the d down to the other side.
So in exp form A^(bd) = C^d so shouldn't the answer be p^(mn) = n rather than P^(q/m) = n ?
If I was given:

$\displaystyle \log_a\left(c^d\right)=bd$

I would first use the identity $\log_a\left(b^c\right)=c\cdot\log_a(b)$ to write:

$\displaystyle d\cdot\log_a\left(c\right)=bd$

Next, divide through by $d$:

$\displaystyle \log_a\left(c\right)=b$

Finally, convert from logarithmic to exponential form:

$\displaystyle c=a^b$